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How To Calculate 99% Confidence Interval

99% Confidence Interval Formula:

\[ CI = \bar{x} \pm t \times \left( \frac{s}{\sqrt{n}} \right) \]

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1. What is a 99% Confidence Interval?

A 99% confidence interval is a range of values that is estimated to contain the true population parameter with 99% confidence. It provides a measure of the uncertainty around a sample statistic.

2. How Does the Calculator Work?

The calculator uses the confidence interval formula:

\[ CI = \bar{x} \pm t \times \left( \frac{s}{\sqrt{n}} \right) \]

Where:

Explanation: The interval is centered around the sample mean, with the width determined by the standard error (s/√n) multiplied by the critical t-value.

3. Importance of Confidence Intervals

Details: Confidence intervals provide more information than point estimates alone, showing the range of plausible values for the population parameter and the precision of the estimate.

4. Using the Calculator

Tips: Enter the sample mean, standard deviation, and sample size. The sample size should be at least 2. For small samples (n < 30), consider using a t-distribution with appropriate degrees of freedom.

5. Frequently Asked Questions (FAQ)

Q1: Why use t=2.576 for 99% CI?
A: This is the critical value from the standard normal distribution (Z) that leaves 0.5% in each tail. For small samples, the t-value would be larger.

Q2: What's the difference between 95% and 99% CI?
A: A 99% CI is wider than a 95% CI, providing greater confidence but less precision. The trade-off is between confidence and interval width.

Q3: When should I use 99% vs 95% CI?
A: Use 99% when you need higher confidence in your interval estimate, accepting a wider range. 95% is more common as it balances confidence and precision.

Q4: What assumptions does this calculation make?
A: Assumes random sampling, approximately normal distribution (especially for small n), and that the sample standard deviation is a good estimate of the population SD.

Q5: How does sample size affect the CI?
A: Larger samples produce narrower CIs (more precise estimates) because the standard error decreases as √n increases.

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